Question

Reflection by thin layers. In Fig. 35-42, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) The waves of rays ${\mathit{r}}_{\mathbf{1}}$and ${\mathit{r}}_{\mathbf{2}}$interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35- 2 refers to the indexes of refraction ${\mathit{n}}_{\mathbf{1}}$, ${\mathit{n}}_{\mathbf{2}}$and${\mathit{n}}_{\mathbf{3}}$, the type of interference, the thin-layer thickness $\mathit{L}$in nanometres, and the wavelength $\mathit{\lambda }$in nanometres of the light as measured in air. Where $\mathit{\lambda }$is missing, give the wavelength that is in the visible range. Where $\mathit{L}$is missing, give the second least thickness or the third least thickness as indicated.

Short answer

The thickness of the thin layer is $329\mathrm{nm}$.

Step-by-step solution

Given data

• The refractive index of first medium is $n_1=1.50$.
• The refractive index of the thin film is $n_2=1.34$
• The refractive index of the third medium $n_3=1.42$
• The maximum intensity occurs at wavelength ${\lambda }_{587}=587nm$.

Interference in thin films

Bright colours reflected from thin oil on water and soap bubbles are a consequence of light interference. Due to the constructive interference of light reflected from the front and back surfaces of the thin film, these bright colours can be seen. For a perpendicular incident beam, the maximum intensity of light from the thin film satisfies the condition:

$2L=\left(m+\frac{1}{2}\right)\frac{\lambda }{n_2}m=0,1,2,...$ (Maxima—bright film in the air)

Where $\lambda$is the wavelength of the light in air, $L$is its thickness, and $n_2$is the film’s refractiveindex.

Determining the 2nd least thickness for this arrangement of materials.

Here, in this case, light travels in medium with $n_1=1.50$and incident on the film whose refractive index is $n_2=1.34$.And then, the light gets reflected of the third surface $n_3=1.42$while travelling through the film.As a result, the condition for constructive interference or maximum intensity is

$2L=\left(m+\frac{1}{2}\right)\frac{{\lambda }_{587}}{n_2}$ (Constructive)

The 2^nd least thickness which $m=1$is

role="math" localid="1663149600515" $\begin{array}{rcl}2L& =& \left(1+\frac{1}{2}\right)\frac{587\mathrm{nm}}{1.34}\\ L& =& \frac{3\left(587\mathrm{nm}\right)}{4\left(1.34\right)}\\ & =& 329\mathrm{nm}\end{array}$

Hence the thickness of the thin layer is $329\mathrm{nm}$.